\(\int x^2 \text {arctanh}(c+d \coth (a+b x)) \, dx\) [298]

Optimal result

Integrand size = 15, antiderivative size = 303 \[ \int x^2 \text {arctanh}(c+d \coth (a+b x)) \, dx=\frac {1}{3} x^3 \text {arctanh}(c+d \coth (a+b x))+\frac {1}{6} x^3 \log \left (1-\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac {1}{6} x^3 \log \left (1-\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac {x^2 \operatorname {PolyLog}\left (2,\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac {x^2 \operatorname {PolyLog}\left (2,\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}-\frac {x \operatorname {PolyLog}\left (3,\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b^2}+\frac {x \operatorname {PolyLog}\left (3,\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b^2}+\frac {\operatorname {PolyLog}\left (4,\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{8 b^3}-\frac {\operatorname {PolyLog}\left (4,\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{8 b^3} \] Output:

1/3*x^3*arctanh(c+d*coth(b*x+a))+1/6*x^3*ln(1-(1-c-d)*exp(2*b*x+2*a)/(1-c+ 
d))-1/6*x^3*ln(1-(1+c+d)*exp(2*b*x+2*a)/(1+c-d))+1/4*x^2*polylog(2,(1-c-d) 
*exp(2*b*x+2*a)/(1-c+d))/b-1/4*x^2*polylog(2,(1+c+d)*exp(2*b*x+2*a)/(1+c-d 
))/b-1/4*x*polylog(3,(1-c-d)*exp(2*b*x+2*a)/(1-c+d))/b^2+1/4*x*polylog(3,( 
1+c+d)*exp(2*b*x+2*a)/(1+c-d))/b^2+1/8*polylog(4,(1-c-d)*exp(2*b*x+2*a)/(1 
-c+d))/b^3-1/8*polylog(4,(1+c+d)*exp(2*b*x+2*a)/(1+c-d))/b^3
 

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 265, normalized size of antiderivative = 0.87 \[ \int x^2 \text {arctanh}(c+d \coth (a+b x)) \, dx=\frac {1}{3} x^3 \text {arctanh}(c+d \coth (a+b x))-\frac {-4 b^3 x^3 \log \left (1+\frac {(1-c+d) e^{-2 (a+b x)}}{-1+c+d}\right )+4 b^3 x^3 \log \left (1+\frac {(-1-c+d) e^{-2 (a+b x)}}{1+c+d}\right )+6 b^2 x^2 \operatorname {PolyLog}\left (2,\frac {(-1+c-d) e^{-2 (a+b x)}}{-1+c+d}\right )-6 b^2 x^2 \operatorname {PolyLog}\left (2,\frac {(1+c-d) e^{-2 (a+b x)}}{1+c+d}\right )+6 b x \operatorname {PolyLog}\left (3,\frac {(-1+c-d) e^{-2 (a+b x)}}{-1+c+d}\right )-6 b x \operatorname {PolyLog}\left (3,\frac {(1+c-d) e^{-2 (a+b x)}}{1+c+d}\right )+3 \operatorname {PolyLog}\left (4,\frac {(-1+c-d) e^{-2 (a+b x)}}{-1+c+d}\right )-3 \operatorname {PolyLog}\left (4,\frac {(1+c-d) e^{-2 (a+b x)}}{1+c+d}\right )}{24 b^3} \] Input:

Integrate[x^2*ArcTanh[c + d*Coth[a + b*x]],x]
 

Output:

(x^3*ArcTanh[c + d*Coth[a + b*x]])/3 - (-4*b^3*x^3*Log[1 + (1 - c + d)/((- 
1 + c + d)*E^(2*(a + b*x)))] + 4*b^3*x^3*Log[1 + (-1 - c + d)/((1 + c + d) 
*E^(2*(a + b*x)))] + 6*b^2*x^2*PolyLog[2, (-1 + c - d)/((-1 + c + d)*E^(2* 
(a + b*x)))] - 6*b^2*x^2*PolyLog[2, (1 + c - d)/((1 + c + d)*E^(2*(a + b*x 
)))] + 6*b*x*PolyLog[3, (-1 + c - d)/((-1 + c + d)*E^(2*(a + b*x)))] - 6*b 
*x*PolyLog[3, (1 + c - d)/((1 + c + d)*E^(2*(a + b*x)))] + 3*PolyLog[4, (- 
1 + c - d)/((-1 + c + d)*E^(2*(a + b*x)))] - 3*PolyLog[4, (1 + c - d)/((1 
+ c + d)*E^(2*(a + b*x)))])/(24*b^3)
 

Rubi [A] (verified)

Time = 1.42 (sec) , antiderivative size = 391, normalized size of antiderivative = 1.29, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6799, 2620, 3011, 7163, 2720, 7143}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int x^2 \text {arctanh}(d \coth (a+b x)+c) \, dx\)

\(\Big \downarrow \) 6799

\(\displaystyle -\frac {1}{3} b (-c-d+1) \int \frac {e^{2 a+2 b x} x^3}{-c-(-c-d+1) e^{2 a+2 b x}+d+1}dx+\frac {1}{3} b (c+d+1) \int \frac {e^{2 a+2 b x} x^3}{c-(c+d+1) e^{2 a+2 b x}-d+1}dx+\frac {1}{3} x^3 \text {arctanh}(d \coth (a+b x)+c)\)

\(\Big \downarrow \) 2620

\(\displaystyle -\frac {1}{3} b (-c-d+1) \left (\frac {3 \int x^2 \log \left (1-\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )dx}{2 b (-c-d+1)}-\frac {x^3 \log \left (1-\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{2 b (-c-d+1)}\right )+\frac {1}{3} b (c+d+1) \left (\frac {3 \int x^2 \log \left (1-\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )dx}{2 b (c+d+1)}-\frac {x^3 \log \left (1-\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{2 b (c+d+1)}\right )+\frac {1}{3} x^3 \text {arctanh}(d \coth (a+b x)+c)\)

\(\Big \downarrow \) 3011

\(\displaystyle -\frac {1}{3} b (-c-d+1) \left (\frac {3 \left (\frac {\int x \operatorname {PolyLog}\left (2,\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )dx}{b}-\frac {x^2 \operatorname {PolyLog}\left (2,\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{2 b}\right )}{2 b (-c-d+1)}-\frac {x^3 \log \left (1-\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{2 b (-c-d+1)}\right )+\frac {1}{3} b (c+d+1) \left (\frac {3 \left (\frac {\int x \operatorname {PolyLog}\left (2,\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )dx}{b}-\frac {x^2 \operatorname {PolyLog}\left (2,\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{2 b}\right )}{2 b (c+d+1)}-\frac {x^3 \log \left (1-\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{2 b (c+d+1)}\right )+\frac {1}{3} x^3 \text {arctanh}(d \coth (a+b x)+c)\)

\(\Big \downarrow \) 7163

\(\displaystyle -\frac {1}{3} b (-c-d+1) \left (\frac {3 \left (\frac {\frac {x \operatorname {PolyLog}\left (3,\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{2 b}-\frac {\int \operatorname {PolyLog}\left (3,\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )dx}{2 b}}{b}-\frac {x^2 \operatorname {PolyLog}\left (2,\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{2 b}\right )}{2 b (-c-d+1)}-\frac {x^3 \log \left (1-\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{2 b (-c-d+1)}\right )+\frac {1}{3} b (c+d+1) \left (\frac {3 \left (\frac {\frac {x \operatorname {PolyLog}\left (3,\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{2 b}-\frac {\int \operatorname {PolyLog}\left (3,\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )dx}{2 b}}{b}-\frac {x^2 \operatorname {PolyLog}\left (2,\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{2 b}\right )}{2 b (c+d+1)}-\frac {x^3 \log \left (1-\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{2 b (c+d+1)}\right )+\frac {1}{3} x^3 \text {arctanh}(d \coth (a+b x)+c)\)

\(\Big \downarrow \) 2720

\(\displaystyle -\frac {1}{3} b (-c-d+1) \left (\frac {3 \left (\frac {\frac {x \operatorname {PolyLog}\left (3,\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{2 b}-\frac {\int e^{-2 a-2 b x} \operatorname {PolyLog}\left (3,\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )de^{2 a+2 b x}}{4 b^2}}{b}-\frac {x^2 \operatorname {PolyLog}\left (2,\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{2 b}\right )}{2 b (-c-d+1)}-\frac {x^3 \log \left (1-\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{2 b (-c-d+1)}\right )+\frac {1}{3} b (c+d+1) \left (\frac {3 \left (\frac {\frac {x \operatorname {PolyLog}\left (3,\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{2 b}-\frac {\int e^{-2 a-2 b x} \operatorname {PolyLog}\left (3,\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )de^{2 a+2 b x}}{4 b^2}}{b}-\frac {x^2 \operatorname {PolyLog}\left (2,\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{2 b}\right )}{2 b (c+d+1)}-\frac {x^3 \log \left (1-\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{2 b (c+d+1)}\right )+\frac {1}{3} x^3 \text {arctanh}(d \coth (a+b x)+c)\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {1}{3} x^3 \text {arctanh}(d \coth (a+b x)+c)-\frac {1}{3} b (-c-d+1) \left (\frac {3 \left (\frac {\frac {x \operatorname {PolyLog}\left (3,\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{2 b}-\frac {\operatorname {PolyLog}\left (4,\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{4 b^2}}{b}-\frac {x^2 \operatorname {PolyLog}\left (2,\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{2 b}\right )}{2 b (-c-d+1)}-\frac {x^3 \log \left (1-\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{2 b (-c-d+1)}\right )+\frac {1}{3} b (c+d+1) \left (\frac {3 \left (\frac {\frac {x \operatorname {PolyLog}\left (3,\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{2 b}-\frac {\operatorname {PolyLog}\left (4,\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{4 b^2}}{b}-\frac {x^2 \operatorname {PolyLog}\left (2,\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{2 b}\right )}{2 b (c+d+1)}-\frac {x^3 \log \left (1-\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{2 b (c+d+1)}\right )\)

Input:

Int[x^2*ArcTanh[c + d*Coth[a + b*x]],x]
 

Output:

(x^3*ArcTanh[c + d*Coth[a + b*x]])/3 - (b*(1 - c - d)*(-1/2*(x^3*Log[1 - ( 
(1 - c - d)*E^(2*a + 2*b*x))/(1 - c + d)])/(b*(1 - c - d)) + (3*(-1/2*(x^2 
*PolyLog[2, ((1 - c - d)*E^(2*a + 2*b*x))/(1 - c + d)])/b + ((x*PolyLog[3, 
 ((1 - c - d)*E^(2*a + 2*b*x))/(1 - c + d)])/(2*b) - PolyLog[4, ((1 - c - 
d)*E^(2*a + 2*b*x))/(1 - c + d)]/(4*b^2))/b))/(2*b*(1 - c - d))))/3 + (b*( 
1 + c + d)*(-1/2*(x^3*Log[1 - ((1 + c + d)*E^(2*a + 2*b*x))/(1 + c - d)])/ 
(b*(1 + c + d)) + (3*(-1/2*(x^2*PolyLog[2, ((1 + c + d)*E^(2*a + 2*b*x))/( 
1 + c - d)])/b + ((x*PolyLog[3, ((1 + c + d)*E^(2*a + 2*b*x))/(1 + c - d)] 
)/(2*b) - PolyLog[4, ((1 + c + d)*E^(2*a + 2*b*x))/(1 + c - d)]/(4*b^2))/b 
))/(2*b*(1 + c + d))))/3
 

Defintions of rubi rules used

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 6799

Int[ArcTanh[(c_.) + Coth[(a_.) + (b_.)*(x_)]*(d_.)]*((e_.) + (f_.)*(x_))^(m 
_.), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcTanh[c + d*Coth[a + b*x]]/(f*( 
m + 1))), x] + (-Simp[b*((1 - c - d)/(f*(m + 1)))   Int[(e + f*x)^(m + 1)*( 
E^(2*a + 2*b*x)/(1 - c + d - (1 - c - d)*E^(2*a + 2*b*x))), x], x] + Simp[b 
*((1 + c + d)/(f*(m + 1)))   Int[(e + f*x)^(m + 1)*(E^(2*a + 2*b*x)/(1 + c 
- d - (1 + c + d)*E^(2*a + 2*b*x))), x], x]) /; FreeQ[{a, b, c, d, e, f}, x 
] && IGtQ[m, 0] && NeQ[(c - d)^2, 1]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

rule 7163

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_. 
)*(x_))))^(p_.)], x_Symbol] :> Simp[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a 
+ b*x)))^p]/(b*c*p*Log[F])), x] - Simp[f*(m/(b*c*p*Log[F]))   Int[(e + f*x) 
^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c 
, d, e, f, n, p}, x] && GtQ[m, 0]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 15.18 (sec) , antiderivative size = 5248, normalized size of antiderivative = 17.32

method	result	size
risch	\(\text {Expression too large to display}\)	\(5248\)

Input:

int(x^2*arctanh(c+d*coth(b*x+a)),x,method=_RETURNVERBOSE)
 

Output:

result too large to display
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 880 vs. \(2 (259) = 518\).

Time = 0.12 (sec) , antiderivative size = 880, normalized size of antiderivative = 2.90 \[ \int x^2 \text {arctanh}(c+d \coth (a+b x)) \, dx=\text {Too large to display} \] Input:

integrate(x^2*arctanh(c+d*coth(b*x+a)),x, algorithm="fricas")
 

Output:

1/6*(b^3*x^3*log(-(d*cosh(b*x + a) + (c + 1)*sinh(b*x + a))/(d*cosh(b*x + 
a) + (c - 1)*sinh(b*x + a))) - 3*b^2*x^2*dilog(sqrt((c + d + 1)/(c - d + 1 
))*(cosh(b*x + a) + sinh(b*x + a))) - 3*b^2*x^2*dilog(-sqrt((c + d + 1)/(c 
 - d + 1))*(cosh(b*x + a) + sinh(b*x + a))) + 3*b^2*x^2*dilog(sqrt((c + d 
- 1)/(c - d - 1))*(cosh(b*x + a) + sinh(b*x + a))) + 3*b^2*x^2*dilog(-sqrt 
((c + d - 1)/(c - d - 1))*(cosh(b*x + a) + sinh(b*x + a))) + a^3*log(2*(c 
+ d + 1)*cosh(b*x + a) + 2*(c + d + 1)*sinh(b*x + a) + 2*(c - d + 1)*sqrt( 
(c + d + 1)/(c - d + 1))) + a^3*log(2*(c + d + 1)*cosh(b*x + a) + 2*(c + d 
 + 1)*sinh(b*x + a) - 2*(c - d + 1)*sqrt((c + d + 1)/(c - d + 1))) - a^3*l 
og(2*(c + d - 1)*cosh(b*x + a) + 2*(c + d - 1)*sinh(b*x + a) + 2*(c - d - 
1)*sqrt((c + d - 1)/(c - d - 1))) - a^3*log(2*(c + d - 1)*cosh(b*x + a) + 
2*(c + d - 1)*sinh(b*x + a) - 2*(c - d - 1)*sqrt((c + d - 1)/(c - d - 1))) 
 + 6*b*x*polylog(3, sqrt((c + d + 1)/(c - d + 1))*(cosh(b*x + a) + sinh(b* 
x + a))) + 6*b*x*polylog(3, -sqrt((c + d + 1)/(c - d + 1))*(cosh(b*x + a) 
+ sinh(b*x + a))) - 6*b*x*polylog(3, sqrt((c + d - 1)/(c - d - 1))*(cosh(b 
*x + a) + sinh(b*x + a))) - 6*b*x*polylog(3, -sqrt((c + d - 1)/(c - d - 1) 
)*(cosh(b*x + a) + sinh(b*x + a))) - (b^3*x^3 + a^3)*log(sqrt((c + d + 1)/ 
(c - d + 1))*(cosh(b*x + a) + sinh(b*x + a)) + 1) - (b^3*x^3 + a^3)*log(-s 
qrt((c + d + 1)/(c - d + 1))*(cosh(b*x + a) + sinh(b*x + a)) + 1) + (b^3*x 
^3 + a^3)*log(sqrt((c + d - 1)/(c - d - 1))*(cosh(b*x + a) + sinh(b*x +...
 

Sympy [F]

\[ \int x^2 \text {arctanh}(c+d \coth (a+b x)) \, dx=\int x^{2} \operatorname {atanh}{\left (c + d \coth {\left (a + b x \right )} \right )}\, dx \] Input:

integrate(x**2*atanh(c+d*coth(b*x+a)),x)
 

Output:

Integral(x**2*atanh(c + d*coth(a + b*x)), x)
 

Maxima [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 277, normalized size of antiderivative = 0.91 \[ \int x^2 \text {arctanh}(c+d \coth (a+b x)) \, dx=\frac {1}{3} \, x^{3} \operatorname {artanh}\left (d \coth \left (b x + a\right ) + c\right ) - \frac {1}{18} \, b d {\left (\frac {4 \, b^{3} x^{3} \log \left (-\frac {{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1} + 1\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (\frac {{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1}\right ) - 6 \, b x {\rm Li}_{3}(\frac {{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1}) + 3 \, {\rm Li}_{4}(\frac {{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1})}{b^{4} d} - \frac {4 \, b^{3} x^{3} \log \left (-\frac {{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1} + 1\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (\frac {{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1}\right ) - 6 \, b x {\rm Li}_{3}(\frac {{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1}) + 3 \, {\rm Li}_{4}(\frac {{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1})}{b^{4} d}\right )} \] Input:

integrate(x^2*arctanh(c+d*coth(b*x+a)),x, algorithm="maxima")
 

Output:

1/3*x^3*arctanh(d*coth(b*x + a) + c) - 1/18*b*d*((4*b^3*x^3*log(-(c + d + 
1)*e^(2*b*x + 2*a)/(c - d + 1) + 1) + 6*b^2*x^2*dilog((c + d + 1)*e^(2*b*x 
 + 2*a)/(c - d + 1)) - 6*b*x*polylog(3, (c + d + 1)*e^(2*b*x + 2*a)/(c - d 
 + 1)) + 3*polylog(4, (c + d + 1)*e^(2*b*x + 2*a)/(c - d + 1)))/(b^4*d) - 
(4*b^3*x^3*log(-(c + d - 1)*e^(2*b*x + 2*a)/(c - d - 1) + 1) + 6*b^2*x^2*d 
ilog((c + d - 1)*e^(2*b*x + 2*a)/(c - d - 1)) - 6*b*x*polylog(3, (c + d - 
1)*e^(2*b*x + 2*a)/(c - d - 1)) + 3*polylog(4, (c + d - 1)*e^(2*b*x + 2*a) 
/(c - d - 1)))/(b^4*d))
 

Giac [F]

\[ \int x^2 \text {arctanh}(c+d \coth (a+b x)) \, dx=\int { x^{2} \operatorname {artanh}\left (d \coth \left (b x + a\right ) + c\right ) \,d x } \] Input:

integrate(x^2*arctanh(c+d*coth(b*x+a)),x, algorithm="giac")
 

Output:

integrate(x^2*arctanh(d*coth(b*x + a) + c), x)
 

Mupad [F(-1)]

Timed out. \[ \int x^2 \text {arctanh}(c+d \coth (a+b x)) \, dx=\int x^2\,\mathrm {atanh}\left (c+d\,\mathrm {coth}\left (a+b\,x\right )\right ) \,d x \] Input:

int(x^2*atanh(c + d*coth(a + b*x)),x)
 

Output:

int(x^2*atanh(c + d*coth(a + b*x)), x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int x^2 \text {arctanh}(c+d \coth (a+b x)) \, dx=\int \mathit {atanh} \left (\coth \left (b x +a \right ) d +c \right ) x^{2}d x \] Input:

int(x^2*atanh(c+d*coth(b*x+a)),x)
 

Output:

int(atanh(coth(a + b*x)*d + c)*x**2,x)